ch-6 linear momentum and impulse-11.pdf

8/10/2019 CH-6 LINEAR MOMENTUM AND IMPULSE-11.pdf

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CHAPTER 6Linear Momentum and Impulse

Units

Momentum and Its Relation to Force Conservation of Momentum Collisions and Impulse Conservation of Energy and Momentum in Collisions Elastic Collisions in One Dimension Inelastic Collisions Collisions in Two or Three Dimensions Center of Mass (CM) Center of Mass and Translational Motion

We make use of the laws of conservation of linear momentum and of energy to analyze collisions.The law of conservation of momentum is particularly useful when dealing with a system of two ormore objects that interact with each other, such as in collisions.

Momentum

The linear momentum p

of an object of mass mmoving with a velocity v

is defined as the

product of the mass and the velocity

p mv=

SI Units are kg m / s Vector quantity, the direction of the momentum is the same as the velocitys

Momentum and Its Relation to Force

p mv=

Momentum is a vector symbolized by the symbol p, and is defined as

pF

t

=

The rate of change of momentum is equal to the net force:

This can be shown using Newtons second law.Momentum is a vector; its direction is the same as the direction of the velocity.

Example 1:For a top player, a tennis ball may leave the racket on the serve with a speed of 55

m/s. If the ball has a mass of 0.060 kg and is in contact with the racket for about 4 ms 34 10x s ,

estimate the average force on the ball. Would this force be large enough to lift a 60-kg person?

2 1p (0.060 )(55 / ) 0 8000.004

mv mv kg m sF N

t t s

= = =

2(60 )(9.80 / ) 600kg m s N

Needed to lift the person. (Yes, enough to lift a person)

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Example 2:Water leaves a hose at a rate of 1.5kg/s witha speed of 20 m/s and is aimed at the side of a car, whichstops it. What is the force exerted by the water on thecar?

p pp f inal initi alF t t

= =

0 30 /30

1.0

k g m sN

s

= =

Conservation of Momentum

During a collision, measurements show that the totalmomentum does not change:

' '

A A B B A A B Bm v m v m v m v+ = +

More formally, the law of conservation of momentum states:

The total momentum of an isolated system of objects remains constant.

Example 3:A 10,000-kg railroad car A, traveling at a speed of 24.0 m/s strikes an identical car B,at rest. If the cars lock together as a result of the collision, what is their common speed justafterward? ' '

A A B B A A B Bm v m v m v m v+ = +

( )A A A B

m v m m v '= +

10,000' (24.0 / )

(10,000 10,000 )

AA

A B

m kgv v

m m kg kg= =

+ +m s

to the right, mutual speed is of car A.

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Example 4:Calculate the recoil velocityof a 5.0-kg rifle that shoots a 0.020-kgbullet at a speed of 620 m/s.

' '

B B R R B B R Rm v m v m v m v+ = +

' '0 0 B B R Rm v m v+ = +

(0.020 )(620 / )

(5.0 )

kg m s

kg=

'' B B

R

R

m vv

m=

2.5 /m s=

Momentum conservation works for a rocketas long as we consider the rocket and its fuelto be one system, and account for the mass

loss of the rocket.

Collisions and Impulse

During a collision, objects are deformeddue to t he large forcesinvolved.

Since , we can wri te

The definition of impulse:

Since the time of the collision is very short, we need not worry about the exact time dependence ofthe force, and can use the average force.

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The impulse tells us that we can get the same change in momentum with a large force acting for ashort time, or a small force acting for a longer time.

This is why you should bend your knees when you land; whyairbags work; and why landing on a pillow hurts less than

landing on concrete.

The air bagincreases the timeof the collision

It will also absorbsome of the energyfrom the body

It will spread out the area of contact decreases the pressure helps prevent penetration wounds

Example 5:A golf ball with a mass or 0.05kg is struck with aclub. The force on the ball varies from zero when contact ismade up to some maximum value and then back to zero whenthe ball leaves the club. If the ball leaves the club at a velocity of+44m/s find the following:

(a) the magnitude of the impulse due to the collision.

p p pf i

I= =

(0.05 )(44 / 0) 2.2 /kg m s kg m s= = +

(b) Estimate the duration of the collision and the average forceacting on the ball.

Displacement =radius of ball about 0.02 m.40.02 9.1 10

22 /

x mt x

v m s

= = = s

3

4

p 2.2 /2.4 10

9.1 10

k g m sF x

t x s

= = = +

N

Conservation of Energy and Momentum in Colli sions

Momentum is conservedin all collisions.Collisions in which kinetic energyis conserved as well are called elasticcollisions, and those in which it is not are called inelastic.

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Elastic Collisions in One Dimension

Here we have two objects colliding elastically. We know the massesand the initial speeds.

Since both momentum and kinetic energy are conserved, we can writetwo equations. This allows us to solve for the two unknown finalspeeds.

Inelastic Collisions

With inelasticcollisions, some of the initial kinetic energy is lost tothermal or potential energy. It may also be gained during explosions,as there is the addition of chemical or nuclear energy.A completely inelasticcollision is one where the objects sticktogetherafterwards, so there is only one final velocity.

Example 6:In a crash test, a car of mass 1500 kg collides with a wall and rebounds. The initial

and final velocities of the car are 15 / and 2.60 /i fv m s v m s= = respectively, if the

collisions lasts for 0.150 s, find:

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(a) the impulse delivered to the car due to the collision.

4p (1500 )( 15.0 / ) 2.25 10 /i i

mv kg m s x kg m s= = =

4p (1500 )( 2.60 / ) 0.390 10 /f fmv kg m s x kg m s= = + = +

4 4p p 0.390 10 / ( 2.25 10 / )f iI x kg m s x kg m s= = + 42.64 10 /I x kg m= s

(b) The size and direction of the force exerted on the car.4

5p 2.64 10 / 1.76 100.150

x kg m sF x

t s

= = = +

N

s

s

llet.f

Example 7:An SUV with mass 1800kg is traveling eastboundat +15.0 m/s, collides with a compact car with mass 900 kgtraveling westbound at -15.0 m/s.

(a) Find the change in the velocity of each car.

1 1 2 2 1 2( )i im v m v m m vf+ = +(1800 )(15.0 / ) (900 )( 15.0 / ) (1800 900 )

fkg m s kg m s kg kg v+ = +

5.00 /f

v m s= +

(b) Find the change in the velocity of each car:

1 1 (5.00 / 15.0 / ) 10.0 /f iv v v m s m s m = = =

2 2 5.00 / ( 15.0 / ) 20.0 /f iv v v m s m s m = = = +

(c) Find the change in the KE of the system consisting of both cars.

2 2 2

1 1 2 2

1 1 1(1800 )(15.0 / )

2 2 2i i iKE m v m v kg m s= + =

2 51 (900 )( 15.0 / ) 3.04 102

kg m s x J + =

2 2

1 2

1 1( ) (1800 900 )(5.00 / )

2 2f fKE m m v kg kg m s= + = +

43.38 10x J=

52.70 10f iKE KE KE x J = =

Example 8:The ballistic pendulum is a

device used to measure the speed of afast-moving projectile such as a buThe bullet is fired into a large block owood suspended from some lightwires. The bullet is stopped by theblock, and the entire system swings upto a height h.

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It is possible to obtain the initial speed of the bullet by measuring h and the two masses.As an example, assume that the mass of the bullet, m1, is 5.00g, the mass of the pendulum, m2, is1.0kg, and h = 5.00cm. Find the initial speed of the bullet, v1.

22 2(9.80 / )(0.05 ) 0.990 /fv gh m s m m= = = s

o

m v m v m m v+ = +

1 1 2 2 1 2( )i i f3

1(5.00 10 ) 0 (1.005 ) fx kg v kg v + =

1 3

(1.005 )(0.990 / )199 /

5.00 10

kg m sv m

x kg

= = s

Collisions in Two or Three Dimensions

Conservation of energy and momentum can also be used t

analyze collisions in two or three dimensions, but unless thesituation is very simple, the math quickly becomesunwieldy.

Here, a moving object collides with an object initially at rest.Knowing the masses and initial velocities is not enough; weneed to know the angles as well in order to find the finalvelocities.

Rocket Propulsion

The operation of a rocket depends on the law of conservation of momentum as applied toa system, where the system is the rocket plus its ejected fuel

This is different than propulsion on the earth where two objects exert forces oneach other

road on car train on track

The rocket is accelerated as a result of the thrust of the exhaust gases This represents the inverse of an inelastic collision

Momentum is conserved Kinetic Energy is increased (at the expense of the stored energy of the rocket fuel)

The initial mass of the rocket is M + m M is the mass of the rocket m is the mass of the fuel

The initial velocity of the rocket is

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The rockets mass is M The mass of the fuel, m, has been ejected The rockets speed has increased to

+

v v

The basic equation for rocket propulsion is:

ln if i ef

Mv v v

M

=

Miis the initial mass of the rocket plus fuel Mfis the final mass of the rocket plus any remaining fuel

The speed of the rocket is proportional to the exhaust speed

Thrust o f a Rocket

The thrust is the force exerted on the rocket by the ejected exhaust gases The instantaneous thrust is given by

e

v MMa M v

t t

= =

The thrust increases as the exhaust speed increases and as the burn rate (M/t)increases

Example 9: A rocket has a total mass of 51 10x kg and a burnout mass of 41 10x kg , including

engines, shell, and payload. The rocket blasts off from Earth and exhausts all its fuel in 4.00 min.

burning the fuel at a steady rate with an exhaust velocity of .34.50 10 /ev x m= s

(a) If air friction and gravity are neglected, what is the speed of the rocket at burnout?

In if i ef

Mv v v

M

= +

53 4

4

1.00 100 (4.5 10 / )In 1.04 10 /1.00 10

x kgx m s x m s

x kg = + =

(b) What thrust does the engine develop at liftoff?6

2 2

5

1.69 10 9.80 / 7.10 /

1.00 10

TT

F x NMa F F Mg a g m s m s

M x kg= = = = =

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(d) Estimate the speed at burnout if gravity isnt neglected.

2 2 3(9.80 / )(2.40 10 ) 2.35 10 /gv g t m s x s x m = = = s

3

4 31.04 10 / 2.35 10 / 8.05 10 /fv x m s x m s x m s= =

Center o f Mass

In (a), the divers motion is pure translation; in (b) itis translationplus rotation. There is one point thatmoves in the samepath aparticle would take ifsubjected to the same force as the diver. This pointis called the center of mass(CM).

The general motion of an object can be considered as the sum of the translational motion of theCM, plus rotational, vibrational, or other forms of motion about the CM.

For two particles, the center of mass lies closer to the one with the most mass:

where Mis the total mass.

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The center of gravity is the point where the gravitational force can beconsidered to act. It is the same as the center of mass as long as thegravitational force does not vary among different parts of the object.

CM for the Human Body

High jumpers have developed a technique where their CMactually passes under the bar as they go over it. This a

them to clear higher bars.

llows

Center of Mass and Translational Motion

This is particularly useful in theanalysis of separations andexplosions; the center of mass(which may not correspond to theposition of any particle) continuesto move according to the netforce.

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CHAPTER 6

LINEAR MOMENTUM

CONCEPTS

1. In a game of pool, the white cue ball hits the #5 ball and stops whilethe #5 ball moves away with the same velocity as the cue ball had

originally. This type of collision is elastic.

2. Kinetic energy is conserved when it is an elastic collision.

3. Two objects move toward each other collide, and separate. If there was no netexternal force acting on the objects, but some kinetic energy was lost, then the

collision was not elastic and total linear momentum was conserved.

4. Two objects collide and bounce off each other. Kinetic energy is conserved

only if the collision is elastic.

5. Kinetic energy is never conserved for a perfectly inelastic collision free of

external forces.

6. Two objects collide and stick together. Kinetic energy is definitely not

conserved.

7. Two objects collide and stick together. Linear momentum is definitely

conserved.

p mv=

8. The product of an objects mass and velocity is equal to momentum.

9. When a cannon fires a cannonball, the cannon will recoil backwardbecause the momentum of the cannonball and cannon is conserved.

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10. A 100-kg football linebacker moving at 2 m/s tackles head-on an

80-kg halfback running 3 m/s. Neglecting the effects due todigging in of cleats, the halfback will drive the linebacker

backwards.

11. A small car meshes with a large truck in a head-on collision.

The small car and large truck experience the same average

force.

12. A small object collides with a large object and sticks. Both objectsexperience the same magnitude of momentum change.

13. When two cars collide and lock together both momentum and total

energy is conserved.

14. A golf ball moving east at a speed of 4 m/s, collides with a stationarybowling ball. The golf ball bounces back to the west, and the bowling

ball moves very slowly to the east. Neither object experiences the

greater magnitude impulse, they are the same.

15. The area under the curve on an F t graph represents impulse.

16. Two equal mass balls, A and B, are dropped from the same height, and rebound

off the floor. The A ball rebounds to a higher position. The A ball is subjected to

the greater magnitude impulse during its collision with the floor.

17. If an object is acted on by a non-zero net external force, its momentum will not remain constant.

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18. A rubber ball and a lump of putty have equal mass. They are thrown with

equal speed against a wall. The ball bounces back with nearly the same speedwith which it hit. The putty sticks to the wall. The ball experiences the greater

momentum. ' 'b b p pbp b pm v m v m vmv+ = +

19. A baseball catcher wears a glove rather than just using bare hands to catch a

pitched baseball because the force on the catchers hand is reduced because

the glove increases the time of impact.

20. In a baseball game, a batter hits a ball for a home run. Compared to the

magnitude of the impulse imparted to the ball, the magnitude of the

impulse imparted to the bat is the same.

21. A freight car moves along a frictionless level railroad track at constant speed.

The car is open on top. A large load of sand is suddenly dumped into the car.This causes the velocity of the car to decrease.

22. For an object on the surface of the earth, the center of gravity and the center of mass arethe same point.

23. Tightrope walkers walk with a long flexible rod in order to lower theircenter of mass.

24. The center of gravity of an object may be thought of as the balance point.

25. A plane, flying horizontally, releases a bomb, which explodes before

hitting the ground. Neglecting air resistance, the center of mass of thebomb fragments, just after the explosion moves along a parabolic path.

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ch-6 linear momentum and impulse-11.pdf

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